Question

Find the directions in the xy-plane in which the following functions have zero change at the given point. Express the directions in terms of unit vectors. $$f(x, y)=12-4 x^{2}-y^{2} ; P(1,2,4)$$

Short answer

Based on the given function and point, determine the directions in the xy-plane in which the function has zero change. The function is $$f(x, y)=12-4 x^{2}-y^{2}$$ and the given point is $$P(1,2,4)$$. The directions in the xy-plane in which the function has zero change at the given point are represented by the vector $$ai-2aj$$, where 'a' can be any scalar value.

Step-by-step solution

Calculate the Gradient Vector

To find the gradient vector, we need to find the partial derivatives of the function with respect to x and y, and then evaluate them at the given point P. The gradient vector is given by the expression $$\nabla f = \begin{bmatrix}\frac{\partial f}{\partial x}\\\frac{\partial f}{\partial y}\end{bmatrix}$$. First, find the partial derivative of f with respect to x, $$\frac{\partial f}{\partial x}= -8x$$ and the partial derivative of f with respect to y, $$\frac{\partial f}{\partial y}= -2y$$ Now, evaluate the partial derivatives at the given point P(1,2,4), $$\frac{\partial f}{\partial x}(1,2) = -8$$ $$\frac{\partial f}{\partial y}(1,2) = -4$$ So, the gradient vector at point P is $$\nabla f(1,2) = \begin{bmatrix}-8 \\ -4 \end{bmatrix}$$

Check when the Directional Derivative is Zero

In the xy-plane, we seek directions where the function has zero change. This is equivalent to finding the directions in which the directional derivative is zero. Let $$\vec{v} = \begin{bmatrix}a \\ b \end{bmatrix}$$ be the direction vector in the xy-plane, then the directional derivative at point P is given by the dot product of the gradient vector and the direction vector: $$D_{\vec{v}} f(1,2) = \nabla f(1,2) \cdot \vec{v} = -8a - 4b$$ Since the function has no change in direction $$\vec{v}$$, its directional derivative must be zero: $$-8a - 4b = 0$$

Simplify the Equation and find the Direction Vector

Since the equation from Step 2 involves only two variables a and b, we can solve for one variable and express it in terms of the other variable. $$-8a - 4b = 0 \Rightarrow b=-2a$$ So, the direction vector in the xy-plane is $$\vec{v} = \begin{bmatrix}a \\ -2a \end{bmatrix}$$

Express the Direction in Terms of Unit Vectors

Now that we have the direction vector, we can express it in terms of unit vectors i and j in the xy-plane, where $$\vec{i}$$ is the unit vector in the x direction and $$\vec{j}$$ is the unit vector in the y direction. The direction vector $$\vec{v} = \begin{bmatrix}a \\ -2a \end{bmatrix}= a\begin{bmatrix}1 \\ -2 \end{bmatrix}=ai-2aj$$ Therefore, the directions in the xy-plane in which the function has zero change at the given point are given by $$ai-2aj$$ where 'a' can be any scalar value.

Key concepts

Gradient Vector

When dealing with multivariable functions, the gradient vector is your best friend. It offers a way to understand how a function changes. The gradient vector, often denoted as \( abla f \), is a vector which consists of the partial derivatives of the function with respect to each variable.
It points in the direction where the function increases the most and its magnitude tells you how fast the function is increasing.

For the function \( f(x, y)=12-4x^2-y^2 \), the gradient at any point (x,y) is obtained by:

• Finding the partial derivatives \( \frac{\partial f}{\partial x} = -8x \) and \( \frac{\partial f}{\partial y} = -2y \).
• The gradient vector here would be \( \begin{bmatrix} -8x \ -2y \end{bmatrix} \).

Calculating the gradient at the point (1, 2) provides \( abla f(1,2) = \begin{bmatrix} -8 \ -4 \end{bmatrix} \). This vector holds all the information about how the function behaves around the point (1,2).

Partial Derivatives

The concept of partial derivatives is fundamental when dealing with functions of multiple variables. Unlike single-variable calculus, where the derivative tells the rate of change along one axis, partial derivatives describe how the function changes as one of the variables changes, while keeping the other constant.
For example:

• The partial derivative of \( f(x, y) = 12 - 4x^2 - y^2 \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = -8x \).
• With respect to \( y \), it's \( \frac{\partial f}{\partial y} = -2y \).

These partial derivatives help in assessing how sensitive the function is to changes in \( x \) and \( y \) individually. They form the building blocks for the gradient vector, enabling the mapping of the function’s behavior in every direction.

Directional Vector

Understanding directional vectors helps us navigate through changes in a function along any path in the plane. These vectors are simply directions defined by a point and directed way. For a function to have zero change in a particular direction at a given point, the dot product of the gradient vector and the directional vector must be zero.

In our case, we found that:

• \( abla f(1,2) = \begin{bmatrix} -8 \ -4 \end{bmatrix} \)
• The directional vector \( \vec{v} = \begin{bmatrix} a \ -2a \end{bmatrix} \)

This directional vector shows that while changing along \( a \), the function remains unchanged. The form \( ai - 2aj \) is a unit vector form in the xy-plane, covering all the directions that ensure no change.

Calculus

Calculus involves the study of how things change. It provides necessary tools, such as derivatives, to explore the behavior of complex functions. In this practice, we apply calculus to explore how a surface, represented by a function, can behave at a point and how it reacts when slightly moved along a plane.

Using calculus:

• We calculated the gradient vector by taking derivative components \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
• Analyzed how the criterion \( -8a - 4b = 0 \) allowed us to determine zero change directions.

Thus, calculus not only gives us these derivative and gradient tools but beautifully illustrates the mapping and potential of multivariable functions.